Given :
Mass of block , m = 1.35 kg .
Speed constant , k = 15 N/m .
Speed at centre , v = 34 cm/s = 0.34 m/s .
To Find :
The amplitude of oscillation .
Solution :
Now , we know total energy is conserved in SHM .
So , K.E at zero displacement :
[tex]K.E=\dfrac{mv^2}{2}\\\\K.E=\dfrac{1.35\times 0.34^2}{2}\\\\K.E=7.8\times 10^{-2}\ J[/tex]
Now , this K.E is equal to maximum P.E :
[tex]P.E=K.E=7.8\times 10^{-2}\ J[/tex] .
[tex]P.E=\dfrac{kA^2}{2}\\\\7.8\times 10^{-2}=\dfrac{kA^2}{2}\\\\A=\sqrt{\dfrac{2\times 7.8 \times 10^{-2}}{15}}\ m\\\\A=0.1019 \ m\\\\A=10.2\ cm[/tex]
Therefore , the amplitude is 10.2 cm .
Hence , this is the required solution .
